⇒Thermodynamics

Correct Option is (C)

Kinetic energy loss of the gas is

$\mathrm{\Delta }\mathrm{E}=\frac{1}{2}M{V}^2\cdot \mathrm{n}$ ..... (i) ....(where $\mathrm{n}$ is the no. of moles of the gas)

Heat gained by the gas due to temperature change $\mathrm{\Delta }\mathrm{T}$ is $\mathrm{\Delta }Q={}_n{C}_V\mathrm{\Delta }T$

But, ${\mathrm{C}}_{\mathrm{V}}=\frac{3}{2}\mathrm{R}$ ....(gas is monoatomic)

$\therefore \mathrm{\Delta }\mathrm{Q}=\frac{3}{2}\mathrm{R}\cdot \mathrm{n}\mathrm{\Delta }\mathrm{T}$ ..... (ii)

Equating (i) and (ii),

$\begin{array}{rl}& \mathrm{\Delta }\mathrm{E}=\mathrm{\Delta }\mathrm{Q}\\ & \frac{1}{2}{\mathrm{MV}}^2\cdot \mathrm{n}=\frac{3}{2}\mathrm{R}\cdot \mathrm{n}\mathrm{\Delta }\mathrm{T}\\ & \mathrm{\Delta }\mathrm{T}=\frac{\mathrm{MV}}{3\mathrm{R}}\end{array}$

Correct Option is (A)

From first law of thermodynamics,

$\mathrm{Q}=\mathrm{\Delta }\mathrm{U}+\mathrm{\Delta }\mathrm{W}=\mathrm{\Delta }\mathrm{U}+\mathrm{P}\mathrm{\Delta }\mathrm{V}$

Change in volume due to compression

$\mathrm{\Delta }V=V_2-V_1=4-10=-6m^3$

Negative sign indicates gas is compressed.

$\therefore \mathrm{\Delta }\mathrm{U}=\mathrm{Q}-\mathrm{P}\mathrm{\Delta }\mathrm{V}=100-[50\times (-6)]=400\mathrm{J}$

As, $\mathrm{\Delta }\mathrm{U}$ is positive, the internal energy is increased.

Correct Option is (D)

Heat supplied at constant pressure ${\mathrm{Q}}_1={\mathrm{nC}}_{\mathrm{p}}\mathrm{dT}$

Heat supplied at constant volume ${\mathrm{Q}}_2={\mathrm{nC}}_{\mathrm{v}}\mathrm{dT}$

Work done

$\begin{array}{rl}& \mathrm{W}={\mathrm{Q}}_1-{\mathrm{Q}}_2=\mathrm{n}({\mathrm{C}}_{\mathrm{p}}-{\mathrm{C}}_{\mathrm{v}})\mathrm{dT}\\ & \therefore \frac{\mathrm{W}}{{\mathrm{Q}}_2}=\frac{{\mathrm{C}}_{\mathrm{p}}-{\mathrm{C}}_{\mathrm{v}}}{{\mathrm{C}}_{\mathrm{v}}}=\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{v}}}-1=\frac{5}{3}-1=\frac{2}{3}\\ & \therefore {\mathrm{Q}}_2=\frac{3\mathrm{W}}{2}\end{array}$